解:(1)在Rt△AOB中,OA=4,OB=3 ∴AB= ①P由O向A运动时,OP=AQ=t,AP=4-t 过Q作QH⊥AP于H点,由QH//BO得
∴ 即 (0<t≤4) ②当4<t≤5时,AP=t-4 AQ=t sin∠BAO= OH= ∴ =··············(4分) (2)由题意知,此时△APQ≌△DPQ ∠AQP=900 ∴cosA= 当0<t≤4 ∴ 即 当4<t≤5时, t=-16(舍去) ∴···············(6分) (3)存在,有以下两种情况 ①若PE//BQ,则等腰梯形PQBE中PQ=BE 过E、P分分别作EM⊥AB于M,PN⊥AB于N 则有BM=QN,由PE//BQ得 ∴ 又∵AP=4-t, ∴AN= ∴由BM=QN,得 ∴ ∴···································(8分) ②若PQ//BE,则等腰梯形PQBE中 BQ=EP且PQ⊥OA于P点 由题意知 ∵OP+AP="OA " ∴ ∴t··············(10分) 由①②得E点坐标为 (4)①当P由O向A运动时,OQ=OP=AQ=t 可得∠QOA=∠QAO ∴∠QOB=∠QBO ∴OQ="BQ=t " ∴BQ=AQ=AE ∴······················(11分) ②当P由A向O运动时,OQ=OP=8-t BQ=5-t, 在Rt△OGQ中,OQ2 =" RG2" + OG2 即(8-t)2 = ∴t = 5·························(12分)
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